Suppose we wanted to evaluate the double integral $S = \iint_D \sqrt{x^2 + y^2} \, dx \, dy$ by first applying a change of variables from $D$ to $R$ : $\begin{aligned} x &= X_1(u, v) = -u^2 + v \\ \\ y &= X_2(u, v) = u^2 + 2u - v \end{aligned}$ What is $S$ under the change of variables? If you know an expression within absolute value is non-negative, do not use absolute value at all. $S = \iint_R $ $du \, dv$
Explanation: If we have a transformation $\bold{X} : R \to D$, then we can rewrite an integral under the change of variables: $ \iint_D f(x, y) \, dA = \iint_R f(\bold{X}(u, v)) | J(\bold{X}) | \, du \, dv$ First we need to find the absolute value of the Jacobian, $|J(\bold{X})|$. $\begin{aligned} |J(\bold{X})| &= \left| \det \begin{pmatrix} \dfrac{\partial X_1}{\partial u} & \dfrac{\partial X_1}{\partial v} \\ \\ \dfrac{\partial X_2}{\partial u} & \dfrac{\partial X_2}{\partial v} \end{pmatrix} \right| \\ \\ &= \left| \det \begin{pmatrix} -2u & 1 \\ \\ 2u + 2 & -1 \end{pmatrix} \right| \\ \\ &= \left| 2u - (2u + 2) \right| \\ \\ &= \left| -2 \right| \\ \\ &= 2 \end{aligned}$ Now we substitute $u$ and $v$ in $f(x, y)$. $\begin{aligned} f(x, y) &= f(\bold{X}(u, v)) \\ \\ &= \sqrt{X_1(u, v)^2 + X_2(u, v)^2} \\ \\ &= \sqrt{ (-u^2 + v)^2 + (u^2 + 2u - v)^2} \end{aligned}$ Putting everything together, we get the integral under the change of variables: $ \iint_R 2 \sqrt{ (-u^2 + v)^2 + (u^2 + 2u - v)^2} \, du \, dv$